How would I go about finding the character table of $C_3$? I know a character of $C_3$ is a homomorphism from $C_3$ to the circle group C. If I take a generator h of $C_3$, elements of $C_3$ are {1, h, $h^2$}.

If I label the character $\phi_0$, $\phi_1$, ..., how does each character interact with each element of $C_3$? Am I looking to first find each character (homomorphism), and then compute $\phi_0$(1), $\phi_0$(h), etc...?

Is the trivial character $\phi_0$ the homomorphism that maps any element of $C_3$ to 1?

Thanks for your help. Sorry I'm quite confused.


Since $C_3$ is simple, every homomorphism is either trivial or an isomorphism onto its image. So aside from the trivial homomorphism, each one sends a generator $h\in C_3$ to a complex number $\zeta$ such that $\zeta^3=1$ and $\zeta^1\ne 1$. Since there are two choices, and each gives a homomorphism, this gives the total character group which is given explicitly as

$$\phi_i:C_3\to S^1:\begin{cases}\phi_0(h)=1 \\ \phi_1(h)=e^{2\pi i\over 3} \\ \phi_2(h) = e^{-2\pi i\over 3}\end{cases}.$$

  • $\begingroup$ Thanks Adam. Could you explain the table at the end of the page of this wolfram link? mathworld.wolfram.com/CyclicGroupC3.html $\endgroup$ – jstnchng Oct 28 '14 at 21:07
  • $\begingroup$ It looks like they have some errors, the table for $C_3$ is not all real. Since any generator has order $3$, it shows that the image always satisfies the polynomial relation $x^3-1=0$, which definitely has properly complex roots. Moreover, it is clear that--aside from the trivial representation which maps everything to $1$, we do not satisfy $x^2-1=0$, since that would imply something is both a first and second root of unity. $\endgroup$ – Adam Hughes Oct 28 '14 at 21:14
  • $\begingroup$ Could you also explain why $C_3$ is simple implies that every homomorphism is either trivial or an isomorphism onto its image? I know that $C_3$ being simple means that $C_3$ does not have any nontrivial normal subgroups. $\endgroup$ – jstnchng Oct 28 '14 at 21:51
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    $\begingroup$ A homomorphism is injective if and only if there is no kernel. Injective things are, by definition, isomorphisms with the image. $\endgroup$ – Adam Hughes Oct 28 '14 at 21:52
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    $\begingroup$ The second part is easy. The third roots of unity are well known. I just picked one to be the image of $h$, then I picked the other one for the other homomorphism. $\endgroup$ – Adam Hughes Oct 28 '14 at 21:53

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